Question

A .25 kg ball initially at rest is hit with a 460 N impact. What is the time of impact for this event if the ball ends up moving 40 m/s

Answer 1

As per Newton's II law we can say

Force applied is rate of change of momentum

`F = (dP)/(dt)`

here we have

`dP = P_f - P_i`

`dP = mv_f - mv_i`

given that

m = 0.25 kg

`v_f = 40 m/s`

`v_i = 0`

F = 460 N

now by above equation

`460 = (0.25(40 - 0))/(t)`

`t = (10)/(460) = 0.022 s`

so it required 0.022 s of contact time